Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $n = \dfrac{6z + 36}{z^2 + 11z + 30} \div \dfrac{7z + 70}{-3z^2 - 15z} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $n = \dfrac{6z + 36}{z^2 + 11z + 30} \times \dfrac{-3z^2 - 15z}{7z + 70} $ First factor the quadratic. $n = \dfrac{6z + 36}{(z + 5)(z + 6)} \times \dfrac{-3z^2 - 15z}{7z + 70} $ Then factor out any other terms. $n = \dfrac{6(z + 6)}{(z + 5)(z + 6)} \times \dfrac{-3z(z + 5)}{7(z + 10)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac{ 6(z + 6) \times -3z(z + 5) } { (z + 5)(z + 6) \times 7(z + 10) } $ $n = \dfrac{ -18z(z + 6)(z + 5)}{ 7(z + 5)(z + 6)(z + 10)} $ Notice that $(z + 6)$ and $(z + 5)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac{ -18z(z + 6)\cancel{(z + 5)}}{ 7\cancel{(z + 5)}(z + 6)(z + 10)} $ We are dividing by $z + 5$ , so $z + 5 \neq 0$ Therefore, $z \neq -5$ $n = \dfrac{ -18z\cancel{(z + 6)}\cancel{(z + 5)}}{ 7\cancel{(z + 5)}\cancel{(z + 6)}(z + 10)} $ We are dividing by $z + 6$ , so $z + 6 \neq 0$ Therefore, $z \neq -6$ $n = \dfrac{-18z}{7(z + 10)} ; \space z \neq -5 ; \space z \neq -6 $